HESI A2 Practice Test Math

Correct Answer: A

Rationale: To solve 4/5 + 6/7, find a common denominator: 35. 4/5 = 28/35 and 6/7 = 30/35. Adding these gives 28/35 + 30/35 = 58/35 = 1 23/35. Therefore, the correct answer is A (1 23/35). The other choices are incorrect as they do not correctly calculate the sum of the fractions. B (2 5/7), C (1 1/7), and D (1 3/4) are not the result of the given expression.

Correct Answer: A

Rationale: The correct answer is A: 3 tablets. To reach the daily intake of 1,500 mg of calcium, the patient should take 1,500 mg ÷ 500 mg per tablet = 3 tablets. Taking 3 tablets will provide the required 1,500 mg of calcium. Choice B (4 tablets) would exceed the required amount, choice C (2 tablets) would be insufficient, and choice D (5 tablets) would exceed the required amount as well. Therefore, A (3 tablets) is the correct answer.

Correct Answer: A

Rationale: Failed to generate a rationale after 5 retries.

Correct Answer: C

Rationale: The correct answer is C: 2700 cu cm. To calculate the volume of a hemispherical bowl, we use the formula V = (2/3)πr³, where r is the radius. Given the diameter of 30cm, the radius is 15cm. Plugging in the values, V = (2/3)π(15)³ = (2/3)π(3375) = 2250π ≈ 7070.78 cu cm. Since the volume of a hemisphere is half of a sphere, we divide by 2 to get 3535.39 cu cm. However, the question asks for the volume of the bowl, not the hemisphere. Since the bowl has an open top, the volume will be double the volume of the hemisphere, giving us 7070.78 cu cm. Therefore, the correct answer is 2700 cu cm (half of the total volume). Choices A, B, and D

Correct Answer: A

Rationale: To find the number of candies each camper will receive, divide the total number of candies (225) by the number of campers (25). 225 ÷ 25 = 9. Therefore, each camper will receive 9 pieces of candy. Choice A is correct because it correctly calculates the equal distribution. Choices B, C, and D are incorrect as they do not divide the total number of candies by the number of campers to determine the equal distribution.